Q.The threshold frequency for a certain metal is $3.3 \times 10^{14}\ \text{Hz}$. If light of frequency $8.2 \times 10^{14}\ \text{Hz}$ is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Concept: Maximum Kinetic Energy — The stopping potential $V_0$ is directly related to the maximum kinetic energy of emitted photoelectrons by $K_{\text{max}} = e V_0$.

Step 1 — Einstein’s photoelectric equation gives:

where $\nu$ is the incident frequency and $\nu_0$ is the threshold frequency.

Step 2 — The cut-off (stopping) voltage $V_0$ satisfies $e V_0 = K_{\text{max}}$, so:

$V_0 = \frac{h}{e}(\nu - \nu_0)$ …

Concept and Intuition

The photoelectric effect tells us that when light of sufficient frequency hits a metal surface, electrons are ejected. The energy of each incoming photon is $h\nu$. Part of this energy is used to overcome the metal's work function $\phi$ (the minimum energy needed to free an electron), and the rest becomes the kinetic energy of the emitted electron.

The maximum kinetic energy of the photoelectrons is given by Einstein's photoelectric equation:

Now, the cut-off voltage (or stopping potential) $V_0$ is the voltage that just stops the most energetic photoelectrons from reaching the other electrode. The work done by this voltage on an electron is $eV_0$, and this must equal the maximum kinetic energy:

So the problem reduces to: find $K_{\text{max}}$ from the given frequencies, then divide by $e$ to get $V_0$.

Step-by-step solution

1. Identify the given data

   Threshold frequency: $\nu_0 = 3.3 \times 10^{14}\ \text{Hz}$

   Incident frequency: $\nu = 8.2 \times 10^{14}\ \text{Hz}$

   Planck's constant: $h = 6.63 \times 10^{-34}\ \text{J·s}$

   Electron charge: $e = 1.6 \times 10^{-19}\ \text{C}$

2. Write the photoelectric equation for stopping potential

   The maximum kinetic energy is:

And since $eV_0 = K_{\text{max}}$, we have:

3. Compute the frequency difference

4. Calculate the numerator $h(\nu - \nu_0)$

First multiply the numbers: $6.63 \times 4.9 = 32.487$ …

Method: Einstein's Photoelectric Equation

This is a direct application of Einstein's photoelectric equation, which connects the maximum kinetic energy of emitted electrons to the frequency of incident light and the work function of the metal.

Step 1: Recall the relevant formula

The maximum kinetic energy of a photoelectron is given by:

where $h$ is Planck's constant ($6.63 \times 10^{-34}\ \text{J·s}$), $f$ is the frequency of incident light, and $\phi$ is the work function of the metal.

The work function is related to the threshold frequency $f_0$ by:

The cut-off voltage $V_0$ (also called stopping potential) is the voltage that just stops the most energetic photoelectrons, so:

where $e = 1.6 \times 10^{-19}\ \text{C}$ is the electron charge.

Step 2: Find the work function

Step 3: Find the maximum kinetic energy

$K_{\text{max}} = 5.4366 \times 10^{-19} - 2.1879 \times 10^{-19}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing cut-off voltage with stopping potential

Students often think cut-off voltage is something separate from stopping potential. It isn't. The cut-off voltage $V_0$ is exactly the stopping potential — the minimum reverse voltage that stops the most energetic photoelectrons.

How to avoid: Remember that the cut-off voltage is defined by $eV_0 = K_{\text{max}}$. When a question asks for cut-off voltage, it is asking for $V_0$ from the photoelectric equation.

Mistake 2: Using the wrong formula or mixing up $f$ and $f_0$

The most common error is plugging numbers into $K_{\text{max}} = hf - \phi$ but forgetting that $\phi = hf_0$. Then students either:

• Use $hf_0$ as the kinetic energy directly, or
• Forget to subtract at all.

How to avoid: Always write the full photoelectric equation step by step:

Then $eV_0 = K_{\text{max}}$, so $V_0 = \frac{h(f - f_0)}{e}$.

Mistake 3: Unit errors with $h$ and $e$

Students use $h = 6.63 \times 10^{-34}\ \text{J s}$ but then forget that $e = 1.6 \times 10^{-19}\ \text{C}$ gives $V_0$ in volts only when $K_{\text{max}}$ is in joules. If you accidentally use $h$ in eV·s, you must adjust accordingly.

How to avoid: Stick to SI units throughout:

• $h = 6.63 \times 10^{-34}\ \text{J s}$
• $e = 1.6 \times 10^{-19}\ \text{C}$
• Frequencies in Hz
• Answer $V_0$ comes out in volts automatically

Mistake 4: Arithmetic errors with powers of 10

When subtracting $3.3 \times 10^{14}$ from $8.2 \times 10^{14}$, students sometimes mess up the exponent or misplace the decimal.

How to avoid: Do the subtraction inside the bracket first:

Then multiply carefully: …